Quotienting a monoid $\Sigma^*$ by a submonoid $A^*$ where $A \subsetneq \Sigma$ are alphabets. I've heard that you can quotient a monoid $M$ and get another monoid $M/\sim$ if you have an equivalence relation $\sim$ on $M$ such that $\forall x,y,z,w \in M$, $x \sim y$ and $z \sim w \implies xz \sim yw$.
But if you have the two Kleene monoids for the alpahabets $A \subsetneq \Sigma$, then you could define on $\Sigma^*$ $x \sim y \iff x \in A^*$ and $y \in A^*$.  But it kind of fulfills the above condition "vacuously", so was wondering if there were another proof that $\Sigma^*$ can be quotiented by $A^*$ whenever $A \subset \Sigma$.
 A: Let $M$ be a monoid and let $P$ be a nonempty subset of $M$. Define a relation $\equiv_P$ on $M$ by setting
$$
x \equiv_P y \quad\text{if and only if}\quad (x \in P \iff y \in P)
$$
This is an equivalence relation of index $2$, since there are exactly two equivalence classes: $(P \times P) \cup (P^c \times P^c)$ and $(P \times P^c) \cup (P^c \times P)$.
Thus if $\equiv_P$ is a congruence, the quotient monoid $M/{\equiv_P}$ has two elements. There are only two monoids of size $2$: the monoid $U_1 = \{0, 1\}$ (with the usual multiplication) and the cyclic group $C_2$ of order $2$.
Let $\pi:M \to M/{\equiv_P}$ be the quotient morphism. If $P$ contains $1$, in particular if $P$ a submonoid of $M$, then $P = \pi^{-1}(1)$. It follows that

A submonoid $P$ of $M$ defines a congruence on $M$ if and only if $P = \pi^{-1}(1)$, where $\pi$ is a surjective morphism from $M$ onto $U_1$ or onto $C_2$.

Coming back to the case of a free monoid $M = A^*$, a submonoid $P$ of $A^*$ defines a congruence if and only if $P = B^*$ (in the $U_1$ case), or, interestingly, if $P = (A^2)^*$ (in the $C_2$ case). Note that $P = (A^3)^*$ does not define a congruence, since in this case, $a \equiv_P a^2$ but $a^2 \not\equiv_P a^3$.
